A variable incline would be something like rolling down the inside of a soup bowl. The local tangent plane at every point increases as you roll up to, say 90 degrees. Conversly, or to say the opposite, the local tangent plane becomes flatter as you roll down to the center of the bowl. As my compadre cyclovenom said, you are missing alot of information in your problem statement.

Disk is rolling (without slipping I assume). So there should be a frictional force F against the motion.

At a time t, say the angle of incline is theta(t). Apply Newton's 2nd law for the motion along the slope (along the tangent at the contact point), and normal to the contact point. Combine these two to get the acceleration of the cm.

But I don't understand why they have given sin(t) instead of sin(theta(t)).

Staff: Mentor

the question asks:a wheel is rolling down variable incline.Show the acceleration of the centre of mass C Acceleration =g[sin t - F/N*Cos t]

I assume you mean:
[tex]a = g[\sin \theta - (F/N) \cos \theta][/tex]

Where theta is the angle of the incline (at the point of contact), F is the friction force (which is also a function of theta), and N is the normal force. (I assume you are using "t" as a shorthand for "theta"?)

This is an odd (but correct) expression for the acceleration. (If the disc is uniform, the acceleration is just a simple function of theta--no need for F and N. But you can certainly write it in terms of F and N, if they insist.)

Apply Newton's 2nd law as Gamma says and you'll get the answer.

The fact that the slope is variable just means the incline has some shape other than flat, as cyrusabdollahi points out. It's irrelevant to this part of the problem, since the acceleration depends only on the slope of the point of contact, which is given by theta.

yes t = theta..couldnt put the theta in the post :$
yes ive tried it..and it has worked. i summed all the forces in the X direction and Y direction and it has worked thanx.

the second part of the question says:Determine an expression for the value of F/N where the only unknowns are the angle (t= theta)and the radius rand the radius of gyration k

i have got as far as
F/N=Tan(t) - a / (g cos t)

i know that I= 1/2 mr^2 for a disc
and that the general formula for I=mk^2
so do i equate the two?..thses are the lines i'm thinking along please tell me if i'm correct or not.. also could you please give me some hints on how i could turn the second term in terms of t,r,k ?

Also, thanx for the help its been driving me mad i really appreciate it.

Staff: Mentor

Quadrophenia said:

i have got as far as
F/N=Tan(t) - a / (g cos t)

OK, that's a reshuffling of the answer from the first part. To eliminate the "a" you'll need another equation: Get it by applying Newton's 2nd law for rotation. (What force exerts a torque on the disc?)

i know that I= 1/2 mr^2 for a disc
and that the general formula for I=mk^2
so do i equate the two?..

Since they want the answer in terms of k, all you need is that last one. (The first one assumes a uniform disc.)

so i did A=F/I which implies that a=Fxr/I using formula a=Ar
m previously worked out was m=N/g cos t
since I=mk^2
my final expression looks like this. (F.r)/(N.k^2) is that right? seems abit weird now come to think of it.:surprised

i dont understand when u say take the torque with respect to the contact point. how does torque come in this. do i use that formula instead.. it tried that and i ended up with a T instead,how do i get rid of it?

Staff: Mentor

Quadrophenia said:

frictional force F=I.A (A= angular acceleration)

As Gamma stated, this is not correct. You need the torque produced by the frictional force.

While you certainly can take torques about the contact point (and that has some advantages) I think it will be more straightforward to find the torque on the disc about its center of mass. The only force exerting such a torque is the friction force F.

Torque is one thing that lets the disc roll here. Do you agree that your formular F = IA is not right? Note that the units in both sides does not match either. That should tell you that this formular is incorrect.

The components of the force along the incline are mg sin@ and F hence the acceleration of C.M. is
a = g sin@ - (F/m)
Components in normal direction gives
N = mg cos@
replace m in from second in first equation.